Problem 43 - Algebra and Trigonometry

Question

Find the intersection points of the pair of ellipses. Sketch the graphs of each pair of equations on the same coordinate axes and label the points of intersection. $$ \left\\{\begin{array}{c}{100 x^{2}+25 y^{2}=100} \\\ {x^{2}+\frac{y^{2}}{9}=1}\end{array}\right. $$

Step-by-Step Solution

Answer

The intersection points are (1, 0) and (-1, 0).

Step 1: Write the Equations in Standard Form

The first ellipse's equation is \( 100x^2 + 25y^2 = 100 \). Divide every term by 100 to transform it into standard form: \( x^2 + \frac{y^2}{4} = 1 \). The second ellipse's equation is already in standard form: \( x^2 + \frac{y^2}{9} = 1 \).

Step 2: Solve the System of Equations Algebraically

Subtract the second equation from the first equation to eliminate \(x^2\): \(\left(x^2 + \frac{y^2}{4} = 1\right) - \left(x^2 + \frac{y^2}{9} = 1\right)\) which leads to \(\frac{y^2}{4} - \frac{y^2}{9} = 0\). Solve this to find \(y^2\).

Step 3: Solve for y

The equation becomes \(\frac{9y^2 - 4y^2}{36} = 0\) or \(\frac{5y^2}{36} = 0\). Therefore, \(y^2 = 0\) or \(y = 0\). Substitute \(y = 0\) into one of the original equations to find \(x\).

Step 4: Substitute y = 0 into the First Equation

Substitute \(y = 0\) into \(x^2 + \frac{y^2}{4} = 1 \) to get \(x^2 + 0 = 1\). Thus, \(x^2 = 1\), which implies \(x = \pm 1\).

Step 5: Identify Intersection Points

The solution \((x, y) = (1, 0)\) and \((-1, 0)\) are the intersection points of the ellipses.

Step 6: Sketch the Ellipses and Mark Intersection Points

Graph the ellipses based on their standard form equations: the first ellipse \(x^2 + \frac{y^2}{4} = 1 \) is a wider ellipse and the second ellipse \(x^2 + \frac{y^2}{9} = 1\) is narrower along the y-axis. Mark the points \((1, 0)\) and \((-1, 0)\) as the intersection points on the graph.

Key Concepts

System of Equations

A system of equations in algebra is a set of equations with multiple variables. The goal is to find values for these variables that satisfy all the equations simultaneously. In our problem, we're working with two equations that describe ellipses:

• \( 100x^2 + 25y^2 = 100 \)
• \( x^2 + \frac{y^2}{9} = 1 \)

These equations describe two ellipses on the same coordinate plane. Our task is to find the intersection points where both ellipses meet. By rewriting both equations in a standard format and using algebraic techniques, we can determine where the graphs of these ellipses intersect. We'll find the specific pairs \((x, y)\) that simultaneously satisfy both equations. This method is essential in determining solutions that are shared between multiple equations.

Standard Form of Ellipses

Ellipses have a standard equation form that helps to identify their size and direction on a coordinate plane. The general standard form of an ellipse is:\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\]Here, \((h, k)\) is the center of the ellipse, while \(a\) and \(b\) are the distances from the center to the ellipse along the x-axis and y-axis, respectively. For our ellipses:

• The first equation \( 100x^2 + 25y^2 = 100 \) is rewritten as \( x^2 + \frac{y^2}{4} = 1 \) in standard form.
• The second equation is already in standard form: \( x^2 + \frac{y^2}{9} = 1 \).

In this form, it's easier to visualize and sketch the ellipses, as well as to understand the relationship between the equations, which will facilitate solving the system algebraically.

Algebraic Solution

An algebraic solution involves manipulating the given equations to find common solutions. In the context of our intersecting ellipses, we're solving the system algebraically to find the intersection points.First, both equations are set into standard form, making it easier to simplify and manipulate. We subtract the second ellipse equation from the first to eliminate one variable, which leads us to: \\(\frac{y^2}{4} - \frac{y^2}{9} = 0\). Solving this, we find \(y^2 = 0\), meaning \(y = 0\).

• Substituting \(y = 0\) into \(x^2 + \frac{y^2}{4} = 1\), results in \(x^2 = 1\).
• This gives us \(x = \pm 1\).

Thus, the intersection points where the ellipses meet are \((1, 0)\) and \((-1, 0)\). This algebraic method is effective in handling systems of equations for geometric shapes like ellipses.